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Chemistry > Topic 12 Acid-Base Equilibria > Flashcards

Topic 12 Acid-Base Equilibria Flashcards

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1
Q

Define an acid.

A

Proton donor.

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2
Q

Define a base.

A

Proton acceptor.

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3
Q

Conjugate acid-base pair

A

A base and its conjugate acid or an acid and its conjugate base.

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4
Q

Conjugate acid

A

When a base accepts a proton, the species formed is its conjugate acid.

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5
Q

Conjugate base.

A

When an acid donates a proton, the species formed is the conjugate base of the acid.

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6
Q

Amphoteric

A

A substance that can act as both an acid and a base.

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7
Q

What substances tend to be bases?

A

They must have a lone pair to accept a proton by forming a dative covalent bond with it. This means a base must contain an atom of the RHS of the periodic table, often oxygen.

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8
Q

What substances tend to be acids?

A

A substance containing a slightly positively-charged hydrogen, e.g., when it is bonded to a highly electronegative atom such as oxygen of a halogen.

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9
Q

Amphiprotic

A

A substance that can both donate and accept protons such as amino acids or HSO4^-. All amphiprotic substances are also amphoteric, but no the other way around.

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10
Q

Monoprotic acids

A

E.g., HCl, can donate one proton. (Also termed monobasic acids.)

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11
Q

Diprotic acids

A

E.g.,H2SO4, can donate 2 protons (also termed dibasic acids).

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12
Q

Diprotic base

A

E.g., a base that can accept 2 protons e.g., CO3^2-+ 2H+ –> H2CO3. Also termed diacidic bases.

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13
Q

Nitric acid as a base in the nitration of benzene.

A

Concentrated H2SO4 + concentrated HNO3 react:
H2SO4 + HNO3<–> HSO4^- + H2NO3^+
H2SO4 is an acid; its conjugate base is HSO4-.
HNO3 is a base; its conjugate acid is H2NO3^+.

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14
Q

When ammonia dissolves in water, this is an example of a conjugate acid-base pair.

A

NH3 (aq) + H2O (l) <–> NH4^+ (aq) + OH^- (aq)
In the forward reaction: NH3 acts as a base & H2O acts as an acid.
In the backward reaction: NH4^+ acts as an acid & OH- acts as a base.

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15
Q

Strong acid

A

Completely dissociates/ionises in aqueous solution.
HCl(aq) –> H+(aq) + Cl-(aq)

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16
Q

Weak acid

A

Only partially dissocisates in aqueous solution, typically organic acids.
CH3COOH(aq) <–> CH3COO-(aq) + H+(aq)

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17
Q

pH of an aqueous solution

A

The reciprocal of the logarithm to the base 10 of the hydrogen ion concentration measured in mol dm^-3. pH= -log[H+]

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18
Q

For strong acids, the [acid] is directly related to the [H+].

A

A solution of HCl of 0.100 mol dm^-3 produces a [H+] of 0.100 mol dm^-3.

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19
Q

How can you determine [H+] from pH?

A

[H+] = 10^-pH

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20
Q

Ka

A

Ka = [H+][A-]/[HA]

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21
Q

[A-]=[H+] at equilibrium. How?

A

Every time a molecule of HA dissociates, a H+ ion & A- ion are formed.

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22
Q

How can we find the pH of a weak acid?

A

Ka= [H+]^2/[HA], assuming the concentration of undissociated acid at equilibrium is equal to the concentration of acid initially as Ka is very small.

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23
Q

pKa

A

-log Ka

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24
Q

The larger the value of Ka…

A

… the stronger the acid.

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25
Q

The smaller the value of pKa…

A

… the stronger the acid.

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26
Q

Sulfuric acid is diprotic and dissociates in 2 stages:

A

H2SO4 (aq) –> H+(aq) + HSO4^-(aq)
HSO4^- (aq) <–> H+ (aq) + SO4^2- (aq)
H2SO4 is a strong acid. HSO4^- is a weak acid.

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27
Q

When determining the pH of H2SO4, why is the contribution to [H+] by HSO4^- negligible?

A

Its dissociation is significantly reduced due to the high [H+] from the full ionisation of H2SO4.

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28
Q

Equation for the self-ionisation of water:

A

H2O (l) <–> H+(aq) + OH^- (aq)

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29
Q

The ionic product of water (Kw)

A

Kw= [H+][OH-]

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30
Q

Kw at 298K

A

1.00 x 10^-14 mol^2 dm^-3

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31
Q

Define a neutral solution.

A

A solution in which [H+] = [OH-]. This is true for pure water.

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32
Q

Wy is pH 7.00 not always neutral?

A

A neutral solution is only pH 7 at 298K as Kw varies with temperature.

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33
Q

Why do even the most alkaline solutions contain some H+ ions?

A

Water ionises.

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34
Q

Why is the small contribution of H+ ions by the water considered insignificant when an acid is dissolved in water?

A

Unless the acid concentration is very small, the acid produces so many H+ ions!

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35
Q

How can we work out the pH of NaOH (strong base) at 0.1 mol dm^-3?

A

The [OH-] = 0.100 mol dm^-3
[H+][OH-] = 1 x 10^-14
[H+]= ( 1 x 10^-14)/0.100= 1 x 10^-13
pH= 13
OR
Use pOH + pH = 14 where pOH= -log[OH-]

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36
Q

Kb

A

Measure of base strength.

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37
Q

Comparing the relative strengths of different acids by pH measurement.

A

Measure the pH of equimolar aqueous solutions of the acids at the same temperature. The higher the value of the pH, the weaker the acid.

38
Q

Comparing the relative strengths of different bases by pH measurement.

A

Measure the pH of equimolar aqueous solutions of the bases at the same temperature. The higher the value of the pH, the stronger the base.

39
Q

Why is the pH of NaCl 7?

A

The salt is made from a strong acid HCl and a strong base NaOH.

40
Q

Why is the pH of KNO3 7?

A

The salt is a product of the strong acid HNO3 and the strong base KOH.

41
Q

Why is an aqueous solution of NH4Cl acidic?

A

The salt is a product of a strong acid (HCl) and a weak base (NH3).

42
Q

Why is an aqueous solution of CH3COONa alkaline?

A

The salt is a product of a weak acid CH3COOH and a strong base NaOH.

43
Q

Why is an aqueous solution of CH3COONH4 neutral?

A

The relative strengths of the weak acid (CH3COOH) and weak base (NH3) that comprise the salt are the same.

44
Q

For each 10-fold decrease in strong acid concentration, what happens to pH?

A

pH increases by 1 unit, until the solution becomes very dilute.

45
Q

For each 10-fold decrease in weak acid concentration, what happens to pH?

A

pH increases by 0.5 units.

46
Q

How to determine Ka of a weak acid experimentally:

A

Weigh 0.50g of weak acid , and dissolve it in 50 cm^3 of deionised water. You may need to warm it for it to dissolve. Transfer the solution to a 250 cm^3 volumetric flask. Add washings from the beaker using deionised water. Make up to the mark using deionised water. Mix the solution by inverting the flask several times. Withdraw a sample, and measure the pH using a calibrated pH meter.

47
Q

Sources of error when determining Ka for a weak acid experimentally.

A
  • We assume [HA] initial = [HA] equilibrium.
  • A small error in the pH reading could cause a large discrepancy in the Ka value.
  • Transfer errors may be significant when weighing small amounts.
48
Q

Endpoint

A

When the indicator just changes colour– not necessarily at the equivalence point.

49
Q

Equivalence point

A

When the acid & base have reacted together in exact proportions as dictated by the stoichiometric equation.

50
Q

Upon what does the pH at the equivalence point depend?

A

The combination of acid & base used.

51
Q

What will the pH be, if you titrate HCl with NaOH?

A

7 as both are strong. The salt NaCl is present at the equivalence point.

52
Q

Titration curve: adding a strong acid to a strong base.

A

pH 7 at the equivalence point. If there is 25 cm^3 of base in the sample, the equivalence point occurs when 25 cm^3 has been added. This is seen as a steep drop in pH.

53
Q

If the acid & base (regardless of strength) react in a 1:1 ratio, when will the equivalence point be on a titration curve, if 25 cm^3 was the amount in the conical flask being titrated?

A

When 25 cm^3 has been added.

54
Q

Titration curve for a weak acid added to a strong base.

A

pH at the equivalence point is 8/9. The curve looks the same as for strong acid + strong base except the vertical region is shorter and past the equivalence point, the acid is present in excess, so the mixture of salt + weak acid forms a buffer solution.

55
Q

Titration curve for a strong acid added to a weak base.

A
  • The pH at the equivalence point is less than 7 as the salt formed is composed of a strong acid + weak base.
  • Before the vertical section & equivalence point, a buffer solution forms containing the weak base & its salt.
56
Q

Titration of a weak acid with a weak base.

A

Difficult to do this titration using an indicator. No vertical section, only a point of inflection at the equivalence point.

57
Q

Acid-base indicator

A

Either a weak acid or a weak base, but usually a weak acid.

58
Q

For an indicator that is a weak acid, how can its dissociation in aqueous solution be represented?

A

HIn (aq) <–> H+ (aq) + In- (aq)
HIn and its conjugate base In- have different colours.

59
Q

Explain how an indicator changes colour according to the pH of the solution using methyl orange as an example.

A

HIn is red. In- is yellow. When [H+] is large enough, the equilibrium shifts left, so that the red colour predominates. If the [H+] is low, the equilibrium lies far to the right, so that the yellow colour predominates.

60
Q

How can the pH at which an indicator changes colour be determined?

A

pH= pKIn

61
Q

When does methyl orange appear orange?

A

[HIn] = [In-]
KIn = [H+][In-]/[HIn]
When [HIn]=[In-], [H+] = KIn
(pH 3.70)

62
Q

How can we calculate the approximate pH range of indicators (example: methyl orange)?

A

The red colour first predominates when [HIn] is 10x [In-]; the yellow colour predominates when [In-] is 10x [HIn].
When [HIn] = 10[In-], and so red first predominates:
[H+][In-]/10[In-] = KIn.
[H+] = KIn x 10
Vice versa to determine when yellow first predominates.
The pH range (between these 2 values) is when methyl orange is orange.

63
Q

The endpoint of a weak acid- weak base titration cannot be determined using an acid-base indicator. What can be used instead?

A

Thermometric titration of a conductometric titration (measuring changes in temperature or electrical conductivity).

64
Q

How can we choose the best indicator?

A

The pH range of the indicator should fall within the vertical section of the graph. Choose an indicator with a pKIn value as close as possible to the equivalence point.

65
Q

Buffer solution

A

A solution that minimises change in pH when a small amount of either acid or base is added.

66
Q

2 of the most common ways to make a buffer solution:

A
  • Mix a weak acid with its conjugate base.
  • Mix a weak base with its conjugate acid.
67
Q

Why are Na & K salts commonly used to make buffers?

A

The salt must be soluble in water.

68
Q

How to calculate the pH of a buffer solution:

A

Mix equal volumes of sodium ethanoate + ethanoic acid, each of 1 mol dm^-3.
We assume the dissociation of CH3COOH is negligible, so CH3COOH at equilibrium= 0.5 mol dm^-3 (it has been halved as the volume has changed).
We assume the concentration of ethanoate ions at equilibrium is due to the sodium salt, so = 0.5.
Ka = [CH3COO-][H+]/[CH3COOH]= 1.74 x 10^-5 mol dm^-3
Rearrange & substitute to find [H+].

69
Q

What happens when a small amount of acid is added to the buffer solution?

A

Most H+ ions react with CH3COO- ions to form CH3COOH.
The total dissociation of CH3COONa means there is a large number of CH3COO- ions, so change in [CH3COO-] is negligible.
The small extent of dissociation of acid means there is a large number of CH3COOH molecules, so change in [CH3COOH] is negligible.
pH remains fairly constant as the [CH3COOH]/[CH3COO-] ratio remains fairly constant, so [H+] remains fairly constant as Ka is constant at a given temperature.

70
Q

What happens when a small amount of base is added to the buffer solution?

A

Most OH- ions react with CH3COOH to form CH3COO- + H2O.
The total dissociation of CH3COONa means there is a large number of CH3COO- ions, so change in [CH3COO-] is negligible.
The small extent of dissociation of acid means there is a large number of CH3COOH molecules, so change in [CH3COOH] is negligible.
pH remains fairly constant as the [CH3COOH]/[CH3COO-] ratio remains fairly constant, so [H+] remains fairly constant as Ka is constant at a given temperature.

71
Q

How can we demonstrate that a buffer solution is resistant to pH changes?

A

Calculate the pH before and after acid/base is added.

72
Q

What is the Henderson-Hasselbalch equation used for?

A

Calculating the pH of a buffer solution.

73
Q

Henderson-Hasselbalch Equation

A

[H+] = Ka x [HA]/[A-]
Taking logs of both sides gives:
pH = pKa - log[acid]/[salt]

74
Q

How to make a buffer solution with a pH less than 7:

A

Use a weak acid + conjugate base.

75
Q

How to make a buffer solution with a pH more than 7:

A

Use a weak base + conjugate acid.

76
Q

Buffer made from a weak base + conjugate acid: ammonia + ammonium ion.

A

NH4^+ <–> NH3 + H+
When acid is added:
H+ + NH3 –> NH4^+
When base is added:
NH4^+ + OH- –> NH3 + H2O
pH= pKa - log[NH4^+]/[NH3]
Thus, pH stays fairly constant as the [NH4^+]/[NH3] ratio stays fairly constant.

77
Q

pH of blood plasma

A

7.35 - 7.45

78
Q

Blood buffer equation

A

CO2(g) + H2O(aq) <–> CO2 (aq) + H2O (aq) <–> H2CO3 (aq) <–> HCO3^-(aq) + H+(aq)

79
Q

Determining Ka from a titration curve.

A

The minimum volume of base required to react completely with the acid is determined from the graph. It is the volume at the equivalence point.
Use the graph to find the pH at the half-equivalence point (when half the volume has been added). This pH= pKa.

80
Q

Why does pH = pKa at the half-equivalence point?

A

The mixture at the half-equivalence point is a buffer solution. the pH is calculated using the H-H equation.
pH = pKa - log[acid]/[salt]
At the half-equivalence point, [salt]=[acid], so [acid]/[salt]=1.
log1= 0
So, pH= pKa.

81
Q

Standard enthalpy change of neutralisation

A

The enthalpy change measured at 100 kPa & 298K when one mole of water is produced by neutralisation of an acid by an alkali.

82
Q

Why is the standard enthalpy change of neutralisation for strong acids + strong bases always roughly the same?

A

Strong acids + strong bases fully dissociate in aqueous solution, so this value is ~ -57.6 kJ mol^-1.

83
Q

Hydration of ions

A

Exothermic!

84
Q

Why is the standard enthalpy change of neutralisation for some weak acids (being neutralised by strong bases) less than 57.6 (e.g., HCN, CH3COOH), but for others it’s greater than 57.6 (e.g., HF)?

A

HA –> H+ + A- Delta H = + x kJ mol^-1
Dissociation= endothermic.
H+(g) + A-(g) –aq–> H+(aq) + A-(aq) Delta H= - y kJ mol^-1
Hydration= exothermic.
When x> y: enthalpy of neutralisation less than 57.6.
when y< x: enthalpy of neutralisation greater than 57.6.

85
Q

How does the buffer system in the blood control pH when extra CO2 is present due to exercise?

A

CO2 dissolves in the blood to form carbonic acid, so the equilibrium shifts right to make more H+. The large reservoir of hydrogen carbonate ions combine with the H+ ions to control pH.

86
Q

Why might an indicator, whose pH range lies partly within the vertical section of the graph, still not be a suitable indicator for a titration?

A

The pH range is not completely within the vertical section. The midpoint of the vertical section (equivalence point) is greater than the midpoint of the pH range of the indicator. This means the colour change would be complete before the equivalence point is reached.

87
Q

Methyl orange

A

Not suitable when weak acids are involved, typically.

88
Q

Why does the pH of a buffer solution scarcely change when a few drops of OH- are added?

A

The OH- reacts with acid to form a salt + water. The pH stays approximately constant because there is a large reservoir of undissociated acid, so the [acid] : [salt] ratio doesn’t change.

89
Q

H3PO4

A

Tribasic acid.

90
Q

Why is water always neutral?

A

[H+] = [OH-] always, even if it’s pH changes.

91
Q

Why is the enthalpy change for the ionisation of water endothermic?

A

Kw increases with temperature.

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